Why does this segment seemingly get added twice in the Rocket Equation? I'm new here, and I'd like to ask a little question about the Tsiolkovsky rocket equation, which I am learning right now.
I am a beginner at this and am just starting to grasp the concept, but there's one point that's getting me confused.
(Please point out any nonsense that I say)
In deriving the rocket equation, we go through this step:
$$mv = (m - \Delta m)(v + \Delta v)$$
which means that the momentum (mass times velocity) of the rocket in the current step (which is AFTER the fuel has been ejected) is [mass of rocket without fuel] times [velocity of rocket after getting the boost from the fuel ejection], from what I can see.
(EDIT: The above paragraph is incorrect; "mv=" does NOT mean that "mv" stands for "momentum after", it means that the "momentum after" is equal to the "momentum before", written as "mv=". "momentum before = momentum after".)
However, it doesn't end there, it extends into:
$$mv = (m-\Delta m)(v+\Delta v)+\Delta m(v-u)$$
$\Delta m (v-u)$ is the momentum of the exhaust gas.
Now, doesn't this mean that they are adding the effect of the exhaust gas twice?
Once in $(v + \Delta v)$, and once in $\Delta m (v-u)$. (since adding the momentum of the gas means that they are adding the "thrusting" from the expulsion of the gas, correct?)
I'm probably misunderstanding something, most likely the interpretation of $(v + \Delta v)$, but I can't wrap my head around it.
Can somebody help?
 A: There are two effects of exhaust gas

*

*accelerating itself away from the spaceship

*accelerating the spaceship.

Because of action reaction law, this will always be present in all interactions. One term for change of motion of one part of the system and one term for change of motion of the other part of the system.

In deriving the rocket equation, we go through this step:
$$mv=(m−\Delta m)(v+\Delta v)$$
which means that the momentum (mass times velocity) of the rocket in the current step (which is AFTER the fuel has been ejected) is [mass of rocket without fuel] times [velocity of rocket after getting the boost from the fuel ejection], from what I can see.

I don't really know what to make of this. Could you provide your source?

$$mv=(m−\Delta m)(v+\Delta v)+\Delta m(v−u)$$

The equation is consequence of conservation of momentum for whole system
$$p_\text{before}=p_\text{after}.$$
Before the ejection, the momentum is simply mass of the rocket before the ejection, times the speed of the rocket before the ejection, i.e.
$$p_\text{before}=m_\text{before} v_\text{before}.$$
After the ejection, the momentum is composed of two parts. One is the momentum of the rocket after the ejection and second one is momentum of ejected fuel, thus the two terms  in your equation, i.e.
$$p_\text{after}=m_\text{rocket after} v_\text{rocket after} + m_\text{fuel after} u_\text{fuel after}.$$
But because of conservation of mass, we have
$$m_\text{rocket after}=m_\text{ before}-m_\text{fuel after}$$
and putting it all together
$$m_\text{before} v_\text{before} = \left(m_\text{before}-m_\text{fuel after}\right) v_\text{rocket after} + m_\text{fuel after} u_\text{fuel after}$$
This leads to your equation, which is then interpreted not as momentum of the rocket after ejection equals these two terms, but as momentum of the rocket before and after ejections are equal, with the momentum after ejection being given by these two terms.
P.S.
To clear notational differences, this is how one arrives at your equation.
$$m_\text{before}=m$$
$$v_\text{before}=v$$
$$m_\text{fuel after}=\Delta m$$
$$v_\text{rocket after}=v+\Delta v$$
$$u_\text{fuel after}=v-u,$$
$u$ being the speed of exhaust gas relative to the rocket and $\Delta v$ the change of speed of the rocket.
In light of @Eli comment

I think the rocket equation is $mv=(m+\Delta m)(v+\Delta v)−\Delta m(v−u)$ thus $dv=−u\frac{dm}{m}$

He interprets his $\Delta m$ as change of mass of the rocket and not mass of the fuel as I did. This introduces negative sign, since the fuel is being expelled and mass of the rocket is being diminished, i.e. it is negative number while  $m_\text{fuel after}$ is positive. This notation is better, because then both $m$ and $\Delta m$ refer to the rocket and not different objects and $\Delta m$ has its usual "change of m" meaning.
A: It's best not to think about "thrust" of the rocket when deriving the rocket equation, but rather think about it in terms of momentum conservation.  The idea here is that an isolated system cannot change its total momentum.
In this case, we consider the system to be the rocket and its fuel.  When a small amount of fuel $\Delta m$ is ejected at a speed $-u$ relative to the motion of the rocket, then the momentum of the system (rocket + fuel) just before the mass is ejected is equal to the momentum of the system (rocket + fuel) just after this fuel is ejected. The second equation you have written represents this statement mathematically:
$$\underbrace{mv}_{\displaystyle {\text{momentum of rocket & fuel}\choose\text{before fuel is ejected}}} = \underbrace{(m-\Delta m)(v+\Delta v)}_{\displaystyle {\text{momentum of rocket}\choose\text{after fuel is ejected}}}+\underbrace{\Delta m(v-u)}_{\displaystyle {\text{momentum of fuel}\choose\text{after it is ejected}}}$$
If we punch the numbers into our calculators, the "before" total on the left-hand side must be equal to the "after" total on the right-hand side.  Note that we must include the momentum of the fuel that was ejected on the right-hand side, since the fuel was part of our system at the start and must remain part of our system at the end.  Omitting this term would amount to ignoring some of the final momentum of the system.
